On Basis Properties of Degenerate Exponential System

Exponential systems of the form are considered, where     0 e n Z x     x  is a degenerate coefficient, is a set of all integers and . The basis properties of these systems in Z 0 Z  Z   π,π L ,1 p p     , when, generally speaking,   x  doesn’t satisfy the Muckenhoupt condition are investigated.


Introduction
Basis properties of classical system of exponents well studied (see e.g.[1][2][3][4]).N. K. Bari in her fundamental work [5] raised the issue of the existence of normalized basis in 2 , which is not Riesz basis.The first example of this was given by K. I. Babenko [6].He proved L that the degenerate system of exponents     .This result has been extended by V. F. Gaposhkin [7].In [8] the condition on the weight  was found, which make the system   Similar problems are considered in [9][10][11][12][13].Basis properties of a degenerate system of exponents are closely related to the similar properties of an ordinary system of exponents in corresponding weight space.In all the mentioned works the authors consider the cases, when the weight or the degenerate coefficient satisfies the Muckenhoupt condition (see, for example, [14]).
In this paper the basis properties of exponential systems with a degenerate coefficient are studied in the spaces , when the degenerate coefficient does not satisfy the Muckenhoupt condition.A similar problem was considered earlier in [15].

Completeness and Minimality
We consider a system of exponents with a degenerate coefficient where   • is a complex conjugate.It is clear that . Then, it follows directly from (2) that 0 and, consequently, Now consider the minimality of system   then it is minimal in and system We have e d e d 2π , , 0 It is clear that in the neighborhood of zero it holds Consequently, the following representation then the system (3) belongs to the space   π, π q L  .Then from the relation (4) we obtain that the system is minimal in in .Let for some function

Since
, then from this relation follows that . As a 0 f  result we obtain that under the following conditions system ( 5) is complete and minimal in .Thus, the system (1) is complete, but it is not minimal in satisfies, then it is known that (see.e.g.[9][10][11][12][13]) system (1) forms basis for    , and in the case 1 p  it is complete and minimal in .Then it is clear that system (5) is minimal, but is not complete in . Now, let the condition (6) holds.It is easy to see that the system , is biorthogonal to the system (5) in .Let us show that in this case the system (5) does not form a basis for At first consider the case 0 Then it is known that (see e.g.[16]) should fulfilled the following conditions 0 0 0 inf sup where p • is an arbitrary norm for .We have Regarding biorthogonal system we get the following condition 0 0 0 inf sup So where p c is a constant depending only on ( in sequel also).Choose p 0   as small as the interval   0, does not contain the points .Then it is absolutely clear that q n p p p q q n q p p q q n q q p p q nt nt 2 It is clear that for sufficiently great we have n And it contradict the condition (7).
Consider the case 0 1 q   .In the absolutely same way as in the previous case, we get Consider the case .We have In the sequel we should pay attention to the following identities From these relations and from the fact that the product of cosines expressed in terms of cosines, it directly follows Theorem 1.Let the following condition be satisfied Then be satisfied.Then the system hold, then system (5) is complete and minimal in , but does not form a basis for it.Proof.If the conditions (8) holds then the system Consequently, the system nZ forms a basis for , and as a result it is complete in   , then as a result it follows that this system does not form a basis for . If the conditions (9) hold then in the absolutely same way as in the previous case we establish that the system (5) is complete and minimal in   1 π, π L  , but does not form a basis for it.Consequently, the system from the previous inequality follows that sup n n I   .Thus, the following theorem is true.

1 .
So the singular operator with the Hilbert kernel is not bounded in